Question

A counter consists of a cylindrical cathode of radius $1cm$ and an anode wire of radius $0.01cm$ which is placed along the axis of the cathode. A voltage of $2.3kV$ is applied between the cathode and anode. The electric field on the anode surface must be :

• A. $2.3\times {10}^{5}V{m}^{-1}$
• B. $5\times {10}^{6}V{m}^{-1}$
• C. $4.6\times {10}^{5}V{m}^{-1}$
• D. $2.5\times {10}^{6}V{m}^{-1}$

Answer 1

Answer: (B)

$5\times {10}^{6}V{m}^{-1}$

Radius of outer cylinder $R_2=1cm$
Radius of inner wire $R_1=0.01cm$
Capacitance of cylindrical capacitor $C=\frac{2\pi {ϵ}_{o}L}{\ln \frac{R_2}{R_1}}$
Given : $V=2300$ volts
So, charge stored on the capacitor $Q=CV=\frac{2\pi {ϵ}_{o}LV}{\ln \frac{R_2}{R_1}}$
Thus charge per unit length $\lambda =\frac{Q}{L}=\frac{2\pi {ϵ}_{o}V}{\ln \frac{R_2}{R_1}}$
Electric field at the anode surface $E=\frac{\lambda }{2\pi {ϵ}_o{R}_1}$
$⟹$ $E=\frac{V}{R_1\ln \frac{R_2}{R_1}}$
Or $E=\frac{2300}{(0.01\times {10}^{-2})\ln 100}=5\times {10}^{6}V/m$